Giáo trình giải tích 2 part 9 potx

x ∈ X d(x, X)=0 X, Y f(x)= d(x, X) d(x, X)+d(x, Y ) . f f −1 (1) = Y, f −1 (0) = X U, V X ⊂ U, Y ⊂ V R n X, Y R n d(X, Y )= inf x∈X,y∈Y d(x, y). K ⊂ R n X K  x → d(x, X) x 0 ∈ K, y 0 ∈ X d(x 0 ,y 0 )=d(K, X) K f : R n → R m B ⊂ R n f(B) f : R n → R m K f −1 (K) f f : K → R K M = {x : f (x)=max K f} [0, 1] (0, 1) f : K −→ f(K) K f −1 K g(x)=sin 1 x (0, +∞) f : A → R m A ⊂ R n f ∃L>0: f(x) − f(y)≤Lx − y, ∀x, y ∈ A f f f : R n −→ R m G f f : C → R C f(x) =0, ∀x ∈ C f(x) x ∈ C x 4 +7x 3 − 9=0 x = x f :[a, b] → [a, b] f x 0 : f(x 0 )=x 0 f [0, 2π] f(0) = f(2π) c ∈ (0, 2π) f (c)=f(c + π) f :[a, b] → R f(a)f(b) < 0 f(x)=0 √ 2 < 1 10 x 2 − 2=0 [0, 2] α ∈ R f(x)=αx, x ∈ R A : R 2 → R 2 A =  ab cd  a, b, c, d > 0 A R 2 + → R 2 + R + = {x ∈ R : x>0} f :[0, π 2 ] → [0, π 2 ] A  cos ϕ sin ϕ  = λ(ϕ)  cos f(ϕ) sin f(ϕ)  f A R 2 + f f : R 2 → R 2 f(x, y)=(ax + by, cx + dy) a, b, c, d f R 2 f : R n → R n f(x)=Ax A =(a ij ) n f :[0,r] → [0,r],f(x)=x 2 r f f : X → X d(f(x),f(y)) <d(x, y), ∀x, y ∈ X,x = y f f :[0, 1] → [0, 1],f(x)=sinx f : K → K K f n = f ◦···◦f    n ln ∩ n∈N f n (K) (f k ) f (x k ) x (f k (x k )) f(x) P k (x)=1+x + ···+ x k k ∈ N f(x)= 1 1 − x 0 <c<1 (P k ) f [0,c] f (0, 1) g [a, b] a = a 0 <a 1 < ···<a n = b g(x)=A k x + B k ,x ∈ [a k−1 ,a k ],k= 1, ···,n A k ,B k g [a, b] B k (f) f(x)=x 2 x ∈ [0, 1] k B k (f) − f =sup x∈[0,1] (|B k (f)(x) − x 2 |) < 1 1000 B k (f) f (x)=x 3 ,x ∈ [0, 1] (B k (f)) f f(x)=e x ,x ∈ R A h(x)= n  i=0 a i e b i x ,n∈ N,a i ,b i ∈ R. f ∈ C[0, 1] A f [a, b] f (P k ) P 0 (x)=0,P k+1 (x)=P k (x)+ 1 2 (x − P k (x) 2 ). 0 ≤ √ x − P k (x) ≤ 2 √ x 1+k √ x 0 ≤ √ x − P k (x) ≤ 2 k . (P k ) [0, 1]  x → √ x f(t)=|t| = √ t 2 ,t ∈ [−1, 1] f ∈ C[0, 1] k =0, 1, ···  1 0 f(x)x k dx =0 f ≡ 0 f  1 0 f 2 =0 f :[0, 1] → R (P k ) f [0, 1] P k P (x) ≤ n n +1 x 0 , ···,x n P (x)= n  i=0 P (x i )π i (x),π i (x)=  j=i (x − x j )  j=i (x i − x j ) ) f : R n −→ R m ∃M f(x)≤M x 2 f 0 Df(0) = 0 f(x) <Mx f f(x, y)=(xy, y/x) f(x, y, z)=(x 4 y, xe z ) f(x, y, z)=(z xy ,x 2 , tgxyz) f(x, y, z)=(e z sin x, xyz) grad f f(x, y, z)=x sin y/z f(x, y, z)=e x 2 +y 2 +z 2 z = x 3 + y 4 x =1,y =3,z =82 x 2 − y 2 + xyz =1 (1, 0, 1) z =  x 2 +2xy −y 2 +1 (1, 1, √ 3) ax 2 +2bxy + cy 2 + dx + ey + f =0 (x 0 ,y 0 ,z 0 ) (2, −1, 2) S 1 : x 2 + y 2 + z 2 =9 S 2 : z = x 2 + y 2 − 3. R 3 S 1 : x 2 + y 2 + z 2 =3 S 2 : x 3 + y 3 + z 3 =3. S 1 ,S 2 (1, 1, 1) R 3 S 1 : ax 2 + by 2 + cz 2 =1 S 2 : xyz =1. a, b, c S 1 ,S 2 c(t)=(3t 2 ,e t ,t+ t 2 ) t =1 c(t)=(2cost, 2sint, t) t = π/2 f(x, y, z)=x 2 y sin z (3, 2, 0) f(x, y)=e x 2 y (0, 0) f (a, b) c f  (c) > 0 f(x)=x x f(x)=sinx x f  (0) > 0 f 0 f [a, b] f  f  (a),f  (b) γ f  (a) f  (b) g(x)=f(x)−γx c ∈ (a, b) f(x)=x 2 sin 1 x x =0 f (0) = 0 f f  (0, 0) f(x, y)= x y , y =0;f(x, y)=0, y =0. f v ∈ R n a D v (a) = lim t→0 f(a + tv) − f(a) t . f a f a D v f(a)=< grad f(a),v > f f f(x, y)= xy x 2 + y x 2 = −y f (x, y)=0 x 2 = −y f(x, y)= x 2 y x 4 + y 2 (x, y) =(0, 0) f(0, 0) = 0 f(x, y)= 3  x 3 + y 3 f(x, y)= xy  x 2 + y 2 x, y =0 f (0, 0) = 0. f(x, y)= x 2 y 2 x 2 y 2 +(y −x) 2 x, y =0 f (0, 0) = 0. f(x, y)=|x| + |y|. f(u, v, w)=u 2 v + v 2 w u = xy, v =sinx, w = e x f(u, v)=u 2 + v sin u u = xe u ,v = yz sin x f : R → R F : R 2 → R F (x, f(x)) ≡ 0 ∂F ∂y =0 f  = − ∂F/∂x ∂F/∂y y = f(x) x = r cos ϕ, y = r sin ϕ f : R 2 → R F (r, ϕ)=f (x, y) (D 1 F (r, ϕ)) 2 + 1 r 2 (D 2 F (r, ϕ)) 2 =(D 1 f(x, y)) 2 +(D 2 f(x, y)) 2 θ (x, y) (u, v) x = u cos θ − v sin θ, y = u sin θ + v cos θ f : R 2 → R F(u, v)=f(x, y) (D 1 F (u, v)) 2 +(D 2 F (u, v)) 2 =(D 1 f(x, y)) 2 +(D 2 f(x, y)) 2 f F (x, y)=f (x 2 + y 2 ) x ∂F ∂y − y ∂F ∂x =0 F (x, y)=f(xy) x ∂F ∂x − y ∂F ∂y =0 F (x, y)=f(ax + by) a ∂F ∂y − b ∂F ∂x =0 f,g : R → R C 2 c ∈ R u(x, y)=f (x + cy) − g(x − cy) u c 2 ∂ 2 u ∂x 2 = ∂ 2 u ∂y 2 v(x, y)=f(3x +2y)+g(x − 2y) 4 ∂ 2 v ∂x 2 − 4 ∂ 2 v ∂x∂y − 3 ∂ 2 v ∂y 2 =0 f : R 2 −→ R 2 ∂f 1 ∂x = ∂f 2 ∂y , ∂f 1 ∂y = − ∂f 2 ∂x . det Jf(x, y)=0 Df(x, y)=0 f f : R n → R m f (tx)=t m f(x), ∀x ∈ R n ,t∈ R + f f m ⇔ n  i=1 x i ∂f ∂x i (x)=mf(x), ∀x ∈ R n . f : R n → R C k (k>1) f(x)=f(0) + n  i=1 g i (x)x i ,g i ∈ C k−1 (R n ). f : R → R |f  (x)|≤L, ∀x f |f(x) − f(y)|≤L| x − y|, ∀x, y ∈ R f : R n → R n f :[a, b] → R 0 <m<f  (x) ≤ M ∀x ∈ [a, b] f(a) < 0 <f(b) f g(x)=x − 1 M f(x) [a, b] x 0 ∈ [a, b] x k+1 = x k − 1 M f(x k ),k ∈ N (x k ) x ∗ f |x k+1 − x ∗ |≤ |f(x 0 )| m  1 − m M  k f : R → R f(a)=b f  (a) =0 δ |x − a| <δ |f  (x) − f  (a)|≤ 1 2 |f  (a)| η = δ 2 |f  (a)| |¯y − b| <η x 0 = a, x k+1 = x k − f(x k ) − ¯y f  (a) (k ∈ N) f(x)=¯y, x ∈ [a −δ, a + δ] f(x, y)=arctgx +arctgy − arctg x + y 1 − xy . f : R n → R m Df(x)=A, ∀x A f f(x)=Ax+ f : U → R U ⊂ R n D 1 f(x)=0, ∀x ∈ U f f(x 1 ,x 2 , ···,x n )=f(x  1 ,x 2 , ···,x n ), ∀(x 1 , ···,x n ), (x  1 , ···,x n ) ∈ U f(x, y)=xy x 2 − y 2 x 2 + y 2 x, y =0 f (0, 0) = 0 ∂ 2 f ∂x∂y (0, 0) = ∂ 2 f ∂y∂x (0, 0). f(x, y)=x 2 + y 2 (0, 0) (1, 2) f(x, y)=e −x 2 −y 2 cos xy (0, 0) f(x, y)=e (x−1) 2 cos y (1, 0) 0 f(x)=e − 1 x 2 x =0 f(0) = 0 f f C 1 u = f 1 (x, y) v = f 2 (x, y) (x 0 ,y 0 ) ∆= ∂f 1 ∂x ∂f 2 ∂y − ∂f 1 ∂y ∂f 2 ∂x 0 (x 0 ,y 0 ) x = x(u, v),y = y(u, v) ∂x ∂u = 1 ∆ ∂v ∂y ∂x ∂v = − 1 ∆ ∂u ∂y ∂y ∂u = − 1 ∆ ∂v ∂x ∂y ∂v = 1 ∆ ∂u ∂x f(x, y)=( x 2 − y 2 x 2 + y 2 , xy x 2 + y 2 ) f (0, 1) R 2  (r, ϕ) → (r cos ϕ, r sin ϕ) ∈ R 2 R 3  (ρ, ϕ, θ) → (ρ cos ϕ sin θ,ρ sin ϕ sin θ, ρcos θ) ∈ R 3 f : R 2 \{(0, 0)}→R 2 \{(0, 0)},f(x, y)=(x 2 − y 2 , 2xy) det Df(x, y) =0, ∀(x, y) f R 2 \ {(0, 0)} f A = {(x, y):x>0} f(A) Dg(1, 0) g f f : R n → R n f(x)=x 2 x f ∈ C ∞ f −1 f(x)= x 2 + x 2 sin 1 x x =0 f(0) = 0 f f  (0) =0 f 0 f : R n → R n C 1 f  (x)≤c<1, ∀x g(x)=x + f(x) g g y (x)=y − f(x) f : R n+k → R n C 1 f(a)=0 Df(a) n c 0 f(x)=c f : R → R C 1 u = f(x) v = −y + xf(x) f  (x 0 ) =0 x 0 ,y 0 ) x = f −1 (u),y= −v + uf −1 (u) x F (x, y)=y 2 + y +3x +1=0 y = y(x) dy dx (x 0 ,y 0 ,z 0 ) z 2 + xy −a =0,z 2 + x 2 − y 2 − b =0. x = f(z),y= g(z) f  (z),g  (z) f : R 3 → R,g: R 2 → R F (x, y)=f(x, y, g(x, y)). DF(x, y) f g F (x, y)=0 x, y D 1 g, D 2 g f  (x 4 + y 4 )/x = u sin x +cosy = v x, y u, v x = π/2,y = π/2 ∂x ∂u (π 3 /4, 1) x, y, z u, v, w (0, 0, 0)      u(x, y, z)=x + xyz v(x, y, z)=y + xy w(x, y, z)=z +2x +3z 2  x 2 − y 2 − u 3 + v 2 +4 = 0 2xy + y 2 − 2u 2 +3v 4 +8 = 0 u, v x, y x =2,y = −1 u(2, −1) = 2,v(2, −1) = 1 u, v u, v x, y  xu + yv 2 =0 xv 3 + y 2 u 6 =0 x =1,y = −1,u =1,v = −1 u = u(x, y),v = v(x, y) u, v, w x, y, z      3x +2y + z 2 + u + v 2 =0 4x +3y + z + u 2 + v + w +2 = 0 x + z + u 2 + w +2 = 0 x =0,y =0,z =0,u=0,v =0,w = −2 u = u(x, y, z),v= v(x, y, z),w = w(x, y, z) sin tx +costx = t, |t| < 1 √ 2 x = ϕ(t) ϕ ϕ 0 Q(x, y)=ax 2 +2bxy + cy 2 (a =0) Q a>0 ac −b 2 > 0 Q a<0 ac − b 2 > 0 Q ac − b 2 < 0 f(x, y)=x 2 +2xy + y 2 +6 f(x, y)=(x 2 + y 2 )e −x 2 −y 2 f(x, y)=x 3 − 3xy 2 f(x, y, z)=x 2 + y 2 +2z 2 + xyz f(x, y, z)=xy 2 z 3 (a − x − 2y −3z) x, y, z > 0 a>0 f(x, y, z)=cos2x sin y + z 2 a 1 , ···,a n ∈ R x n  i=1 (x − a i ) 2 min n x, y x x 1 x 2 ··· x m y y 1 y 2 ··· y m n p(x)=a 0 + a 1 x + ···+ a n x n Q(a 0 , ···,a n )= m  i=1 (p(x i ) − y i ) 2 → min p(x)=a 0 + a 1 x  a 1  i x 2 i + a 0  i x i =  i x i y i a 1  i x i + na 0 =  i y i x y x y f :[0, 1] → R A, B, C  1 0 (f(x) − Ax 2 − Bx − C) 2 dx min A, B, C                1 5 A + 1 4 B + 1 3 C =  1 0 x 2 f(x)dx 1 4 A + 1 3 B + 1 2 C =  1 0 xf(x)dx 1 3 A + 1 2 B + C =  1 0 f(x)dx Ax 2 + Bx + C 2 f n [a, b] . 1, √ 3) ax 2 +2bxy + cy 2 + dx + ey + f =0 (x 0 ,y 0 ,z 0 ) (2, −1, 2) S 1 : x 2 + y 2 + z 2 =9 S 2 : z = x 2 + y 2 − 3. R 3 S 1 : x 2 + y 2 + z 2 =3 S 2 : x 3 + y 3 + z 3 =3. S 1 ,S 2 (1, 1, 1) R 3 S 1 :. y)=ax 2 +2bxy + cy 2 (a =0) Q a>0 ac −b 2 > 0 Q a<0 ac − b 2 > 0 Q ac − b 2 < 0 f(x, y)=x 2 +2xy + y 2 +6 f(x, y)=(x 2 + y 2 )e −x 2 −y 2 f(x, y)=x 3 − 3xy 2 f(x, y, z)=x 2 + y 2 +2z 2 +. cy) − g(x − cy) u c 2 ∂ 2 u ∂x 2 = ∂ 2 u ∂y 2 v(x, y)=f(3x +2y)+g(x − 2y) 4 ∂ 2 v ∂x 2 − 4 ∂ 2 v ∂x∂y − 3 ∂ 2 v ∂y 2 =0 f : R 2 −→ R 2 ∂f 1 ∂x = ∂f 2 ∂y , ∂f 1 ∂y = − ∂f 2 ∂x . det Jf(x, y)=0